A vector\vec{A} is rotated by a small angle \Delta \Theta radians\left ( \Delta \Theta < < 1 \right )    to get a new vector \vec{B}\left .In that case  \left | \vec{B}-\vec{A} \right |  is :

  • Option 1)

    0

  • Option 2)

    \left | \vec{A} \right |\left ( 1-\frac{\Delta \Theta ^{2}}{2} \right )

  • Option 3)

    \left | \vec{A} \right |\Delta \Theta

  • Option 4)

    \left | \vec{B} \right |\Delta \Theta\left -| \vec{A} \right |

 

Answers (1)

As we discussed in

Triangle law of vector Addition -

If two vector are represented by both magnitude and direction by two sides of triangle taken in same order then their resultant is represented by 3rd side of triangle. 

- wherein

Represents triangle law of vector Addition

 

 By Triangle Rule

\vec{A}+\vec{C}= \vec{B}

\vec{B}-\vec{A}= \vec{C}

\left |\vec{B}-\vec{A} \right |=\left | \vec{C} \right | =\left | \vec{B} \right |\sin \Theta (\because \Delta \Theta < < 1)

\left | \vec{B}-\vec{A} \right |=\left | \vec{B}\right | \Delta\Theta\ \; (\because sin\Delta \Theta =\Delta \Theta)

Again \left | \vec{B}\right |cos\Delta \Theta =\left | \vec{A}\right |

\left | \vec{B}\right | =\left | \vec{A}\right |             (\because cos\Delta \Theta \simeq 1)

So \left | \vec{B-A}\right | =\left | \vec{B}\right |\Delta \Theta =\left | \vec{A}\right |.\Delta \Theta

 

 


Option 1)

0

Incorrect

Option 2)

\left | \vec{A} \right |\left ( 1-\frac{\Delta \Theta ^{2}}{2} \right )

Incorrect

Option 3)

\left | \vec{A} \right |\Delta \Theta

Correct

Option 4)

\left | \vec{B} \right |\Delta \Theta\left -| \vec{A} \right |

Incorrect

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